Question

Two cruise ships leave the same port with a 35° angle between their path. Cruise A is traveling at 18 miles per hour and Cruise B is traveling at 15 miles per hour. If they travel in a straight path, find the distance between the cruise ships after 2 hours

Answer 1

9514 1404 393

Answer:

20.7 miles

Explanation:

The distance that A travels in the same direction as B is ...

(18 mi/h)(2 h)cos(35°) = 29.489 mi

So, the difference in distances in that direction is ...

(15 mi/h)(2 h) -29.489 mi = 0.511 mi

__

The distance A travels in the direction perpendicular to B is ...

(18 mi/h)(2 h)sin(35°) = 20.649 mi

So, the straight-line distance between the ships is the hypotenuse of the right triangle with these distances as legs:

AB = √(0.511² +20.649²) = √426.632 . . . miles

AB = 20.655 miles ≈ 20.7 miles . . . separation after 2 hours

Answer 2

`20\:\mathrm{miles}`

Explanation:

After travelling two hours, the two cruise ships form a triangle. One of the legs of this triangle will be the distance Cruise A travelled, and another will be the distance Cruise B travelled. We can find the distance they travel using:

`s=(d)/(t), d=s\cdot t`

Cruise A is travelling at 18 miles per hour for 2 hours. Therefore, Cruise A has travelled:

`d=s\cdot t=18\cdot 2=36\:\mathrm{miles}`

Cruise B is travelling at 15 miles per hour for 2 hours. Therefore, Cruise B has travelled:

`d=s\cdot t=15\cdot 2=30\:\mathrm{miles}`

Because we are given the angle between these legs, we can use the Law of Cosines to find the third leg. The Law of Cosines is given by:

`c^2=a^2+b^2-2ab\cos C`, where
`a`,
`b`, and
`c` are interchangeable. Let
`c` represent the distance between the two cruises. We have:

`c^2=36^2+30^2-2\cdot36\cdot30\cdot \cos 35^(\circ),\\c^2=426.63,\\c\approx \fbox{$20\:\mathrm{miles}$}`(one significant figure).